I happen to have an explicit calculation written down, so here goes.
The key result to be familiar with is the fundamental theorem of symmetric polynomials. By this I mean not the result but with the process of converting any symmetric polynomial into expressions in the elementary symmetric polynomials. In our case, the coefficients of the given cubic are $-\sigma_1$, $\sigma_2$, and $-\sigma_3$, via the expansion of $(X-x_1)(X-x_2)(X-x_3)$.
I'm afraid that this will get (even more) unwieldy unless I condense notation. I will use $\Sigma x_1^ax_2^bx_3^c$ to refer to the sum of the elements in the orbit of the monomial $x_1^ax_2^bx_3^c$ under the action of the symmetric group. For instance $\Sigma x_1 = x_1 + x_2 + x_3 = \sigma_1$ and $\Sigma x_1^3x_2^2x_3 = x_1^3x_2^2x_3+x_1^3x_2x_3^2+x_1^2x_2^3x_3+x_1^2x_2x_3^3+x_1x_2^3x_3^2+x_1x_2^2x_3^3$.
First, I expand $\Delta$, which is not symmetric:
$\Delta = (x_1-x_2)(x_1-x_3)(x_2-x_3) = x_1^2x_2 - x_1^2x_3 - x_2^2x_1 + x_2^2x_3 - x_3^3x_2 +
x_3^2x_1$.
Now we square it! Using the condensed notation I check directly
$\Delta^2 = \Sigma x_1^4x_2^2 - 2\Sigma x_1^4x_2x_3 - 2
\Sigma x_1^3x_2^3 +2\Sigma x_1^3x_2^2x_3 - 6
x_1^2x_2^2x_3^2$.
Now, it only remains to write this symmetric polynomial as a polynomial in $\sigma_1, \sigma_2, \sigma_3$. This is pretty tedious, but entirely mechanical. The leading terms justify why I'm choosing these particular expansions. As an aside, I compute only the coefficients of one monoial per orbit, and I also plug in $x_1=x_2=x_3=1$ as a double-check.
- $\sigma_1^2\sigma_2^2 = \Sigma x_1^4x_2^2 + 2\Sigma
x_1^4x_2x_3 + 2\Sigma x_1^3x_2^3 +8\Sigma
x_1^3x_2^2x_3+ 15x_1^2x_2^2x_3^2$
$\sigma_1^3\sigma_3 = \Sigma x_1^4x_2x_3 + 3\Sigma x_1^3x_2^2x_3 + 6
x_1^2x_2^2x_3^2$
$\sigma_2^3 = \Sigma x_1^3x_2^3 + 3\Sigma x_1^3x_2^2x_3 +
6x_1^2x_2^2x_3^2$
$\sigma_1\sigma_2\sigma_3 = \Sigma x_1^3x_2^2x_3 + 3x_1^2x_2^2x_3^2$
$\sigma_3^2 = x_1^2x_2^2x_3^2$
Now, it's just a matter of finding the right linear combination
- $\Delta^2 = \sigma_1^2\sigma_2^2 - 4\sigma_1^3\sigma_3 - 4\sigma_2^3
+ 18\sigma_1\sigma_2\sigma_3 - 27 \sigma_3^2$.
Replace $\sigma_1,\sigma_2, \sigma_3$, with $-b, c, -d$, and you have the discriminant of $x^3+bx^2+cx+d$.
You can use Vieta's formulas. In particular, with an initial coefficient of $1$, i.e.,
$$P(s) = s^3 + a_2 s^2 + a_1 s + a_0 \tag{1}\label{eq1}$$
has $a_0 = -xyz = -18$, $a_1 = xy+yz+xz = -6$ and $a_2 = -(x + y + z) = 3$. Thus, you have
$$P(s) = s^2 + 3s^2 - 6s - 18 \tag{2}\label{eq2}$$
You could use this to solve for the individual roots using the Cubic function formula.
Update: As discussed in the question comments and the comments below, it's often easier & faster, especially for simpler type cubic equations (e.g., all coefficients are integers) to first try other methods to determine one root, e.g., by factoring by grouping (see How to Factor by Grouping for details) or using the Rational root theorem. If you do determine a root, you can then reduce your sets of equations to a quadratic, so you can then use the Quadratic formula to get the other $2$ roots. If you can't find an initial root, note the suggestion above of using the cubic function formula will always work.
Best Answer
[tl;dr]$\;$ The resulting sextic is solvable, but the calculations are extremely laborious, so for all practical purposes the answer would be to solve it numerically.Let the general depressed cubic be $P(x) = x^3 + a x + b$ then the period-$2$ periodic points of $P$ other than the fixed points are the roots of the sextic:
$$ \frac{P\left(P(x)\right) - x}{P(x) - x} = x^6 + (2 a + 1) x^4 + 2 b x^3 + (a^2 + a + 1) x^2 + (2 a + 1) b x + a + b^2 + 1 $$
The Tschirnhaus transformation $\,z = x^3 + (a - 1) x + b\,$ reduces it to the cubic in $t = z^2\,$:
$$ t^3 + 6 a \,t^2 + 3 (3 a^2 - 4 a - 4) \,t + 4 a^3 - 12 a^2 + 27 b^2 + 16 $$
It follows that the original sextic is solvable.
It would be technically possible to solve the latter cubic, then for each of the three $t$ roots calculate the corresponding $z = \pm t^{1/2}$ and solve each of the six Tschirnhaus cubics. That would give $18$ potential $x$ roots for the original sextic, out of which the $6$ actual roots would need to be identified and retained. Or, once established that the sextic is in fact solvable, follow the reference quoted on wikipedia's page on sextics T. R. Hagedorn, General formulas for solving solvable sextic equations, J. Algebra 233 (2000), 704-757, to the tune of 54 loaded pages. Either way, the exact calculations are very heavy, so numerical root-finding would be preferred in practical applications.
[ EDIT ] $\;$ Finding the Tschirnhaus transformation was essentially trial and error. I thought I'd look for a bicubic (if that's what a cubic in $z^2$ can be called) and started with a depressed cubic substitution $z = x^3 + u x + v$. Canceling the $x^5$ term gave $v = b$ and canceling the $x^3$ term required $u \in \{a-1, a+1/2, a+2\}$. Out of the three choices, $u=a-1$ turned out to cancel the linear term as well.
[ EDIT #2 ] $\;$ Another Tschirnhaus transformation that settles the question of solvability even more directly is $z = x^3 + (a+1)x +b$, which reduces the sextic to (the square of) a cubic:
$$ z^3+(a+2)z-b $$
In light of the related question Irreducibility and Galois group of the 2 -periodic points of a (cubic) polynomial it may be worth noting that something similar holds true for the reduced quartic $P(x)=x^4+ax+b$. In that case $\frac{P\left(P(x)\right)-x}{P(x)-x}$ has degree $12$, and the substitution $z=P(x)+x$ reduces it to (the square of) a sextic.